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(A) According to the law of conservation of mechanical energy,the initial kinetic energy of the block is converted into the elastic potential energy o

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$\sqrt{Mk} \, L$

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(A) According to the law of conservation of mechanical energy,the initial kinetic energy of the block is converted into the elastic potential energy of the spring at the point of maximum compression.Let $v$ be the velocity of the block just before it hits the spring.$\frac{1}{2} M v^2 = \frac{1}{2} k L^2$Solving for $v$:$v^2 = \frac{k L^2}{M} \implies v = L \sqrt{\frac{k}{M}}$The momentum $p$ of the block is given by $p = Mv$.Substituting the value of $v$:$p = M \left( L \sqrt{\frac{k}{M}} \right) = L \sqrt{M^2 \cdot \frac{k}{M}} = L \sqrt{Mk} = \sqrt{Mk} \, L$Thus,the maximum momentum of the block is $\sqrt{Mk} \, L$.

$A$ block of mass $M$ moves on a frictionless horizontal surface and collides with a spring of spring constant $k$. The spring is compressed by a length $L$. What is the maximum momentum of the block after the collision?

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